Two rods of lengths $a$ and $b$ slide along the coordinate axes which are rectangular in such a manner that their ends are concyclic. Then, the locus of the centre of the circle is:

x^{2} + y^{2} = a^{2} + b^{2}

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x^{2} - y^{2} = 4 (a^{2} - b^{2})

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4 (x^{2} - y^{2}) = a^{2} - b^{2}

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x^{2} - y^{2} = a^{2} - b^{2}

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Solution

The correct option is C $4 (x^{2} - y^{2}) = a^{2} - b^{2}$
Length of intercept made by the circle on x-axis $= 2 \sqrt{g^{2} - c}$
Length of intercept made by the circle on y-axis $= 2 \sqrt{f^{2} - c}$
$a = A B = 2 \sqrt{g^{2} - c}$
$b = A B = 2 \sqrt{f^{2} - c}$
Let the locus be P(x,y)
$a^{2} - b^{2} = 4 (x^{2} - c) - 4 (y^{2} - c)$
$= 4 (x^{2} - y^{2})$
$a^{2} - b^{2} = 4 (x^{2} - y^{2})$

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